An artillery shell is fired from $x = 0$ so as to land in one piece at $x = 1000 m$ . Instead it explodes at the top of its parabolic trajectory into two pieces of equal mass. The first piece lands at $x = 750 m$ . The other piece happens to hit the ground at the same instant. It lands at $x =$ (in $m$ )

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Solution

Given range of the shell is $= 1000 m$ ,
As there is no external force during collision,
COM of broken pieces together is same as that of COM of whole piece without explosion
$X_{cm} = \frac{M_{1} x_{1} + M_{2} x_{2}}{M_{1} + M_{2}}$
$1000 = \frac{\frac{M}{2} (750) + \frac{M}{2} (x)}{M}$
$2 (1000 - 375) = x$
$x = 1250 m$

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An artillery shell is fired from x=0 so as to land in one piece at x=1000 m. Instead it explodes at the top of its parabolic trajectory into two pieces of equal mass. The first piece lands at x=750 m. The other piece happens to hit the ground at the same instant. It lands at x= (in m)

Given range of the shell is =1000 m, As there is no external force during collision, COM of broken pieces together is same as that of COM of whole piece without explosion Xcm=M1x1+M2x2M1+M2 1000=M2(750)+M2(x)M 2(1000−375)=x x=1250 m